Given a set $E \subset \mathbb{R}^d$, then define the $\delta-$packing of $E$ as $$\mathcal{P}(r,E)=\{ \{B_i \}_{i\geq 1} ; B_i \text{ is a ball centered on } E, \text{the radius of } B_i \text{ is } r \text{ for any } i, \text{and for any } i \neq j, B_i \cap B_j = \emptyset \}.$$ Then we define $N_{r}'(E)$ as the largest number of balls of radius $r$ needed for the packing of set $E$ . For simplicity, we denote $N_{R,r}'(E)$ as $\inf_{x \in E} N_{r}'(B(x,R) \cap E).$
My question is , for a given set $E \subset \mathbb{R}^d$, and for any $r_3 < r_2 <r_1$, is the following inequation $$N'_{r_3}({B(x,r_1)\cap E}) \geq N'_{r_2}({B(x,r_1)\cap E}) \cdot \inf_{y\in E} N'_{r_3}({B(y,r_2)\cap E})$$ true for the given $x \in E?$ More generally, is the inequation true without the choice of $x \in E$, that is, is the inequation $$N'_{r_1,r_3}( E ) \geq N'_{r_1,r_2}(E) \cdot N'_{r_2,r_3}(E)$$ true?
It is not true in general. Consider the interval $E=[0,4]\subset\mathbb{R}$, and radii \begin{align} r_1&=4\\ r_2&=2(1-\epsilon)\\ r_3&=1(1-2\epsilon) \end{align} with $\epsilon>0$ small, say $0.01$. Then you can check that $$N'_{r_1,r_3}=N'_{r_3}(B(x,r_1)\cap E)=3$$ $$N'_{r_2,r_3}=N'_{r_2}(B(x,r_1)\cap E)=2$$ for every $x\in E$, and finally $$N'_{r_2,r_3}=N'_{r_3}(B(x,r_2)\cap E)=2$$ for $x$ outside a small interval around $\{2\}$, for instance $x=4$ is ok. In particular the inequality is not satisfied.